Higher January 2021 Paper 1R Q21
21 The \(n\)th term of an arithmetic series is \(u_n\) where \(u_n \gt 0\) for all \(n\)
The sum to \(n\) terms of the series is \(S_n\)
Given that \(u_4 = 6\) and that \(S_{11} = (u_6)^2 + 18\)
find the value of \(u_{20}\)
(6)
| Scheme | Marks |
|---|---|
| E.g. \(a + 3d = 6\) oe | M1 |
E.g. \(\dfrac{11}{2}(2a + 10d) = (a + 5d)^2 + 18\) | M1 |
E.g. \(a = 6 - 3d\) and \(\dfrac{11}{2}\left[2(6 - 3d) + 10d\right] = (6 - 3d + 5d)^2 + 18\) or \(d = \dfrac{6 - a}{3}\) and \(\dfrac{11}{2}\left[2a + 10\left(\dfrac{6 - a}{3}\right)\right] = \left(a + 5\left(\dfrac{6 - a}{3}\right)\right)^2 + 18\) | M1 |
| E.g. \(2d^2 + d - 6\;(= 0)\) oe or \(2a^2 - 27a + 36\;(= 0)\) oe | A1 |
| \(d = 1.5\) oe and \(a = 1.5\) oe | A1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 30 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for forming an equation for the 4th term of the sequence
M1: for forming an equation for the sum of the first 11 terms of the sequence
M1: dep on M2 for a correct first step to solve the two equations (writing the equation in terms of one variable)
Note:
If \(\dfrac{11}{2}(2a + 10d) = (a + 5d)^2 + 18\) is expanded then this must be a correct expansion
E.g.
\(11a + 55d = a^2 + 10ad + 25d^2 + 18\)
A1: for a correct 3 term quadratic equation
A1: for a correct value of \(d\) and \(a\)
A1: cao